Using division, we get
\frac{2x^3-3x^2-4x-3}{-5x^2+2x-4}=-2/5x+\frac{11x^2+28x+15}{5(5x^2-2x+4)}
=-2/5x+11/25+81/125\frac{10x-2}{5x^2-2x+4}+\frac{317}{125}\frac1{5x^2-2x+4}
\therefore \int \frac{2x^3-3x^2-4x-3}{-5x^2+2x-4}\ dx
=\int(-2/5x+11/25+81/125\frac{10x-2}{5x^2-2x+4}+\frac{317}{125}\frac1{5x^2-2x+4})dx
=-2/5\intx\ dx+11/25\int dx+81/125\int \frac{(10x-2)dx}{5x^2-2x+4}+\frac{317}{125}\int \frac{dx}{5x^2-2x+4}
=-2/5\frac{x^2}{2}+11/25x+81/125\int \frac{d(5x^2-2x+4)}{5x^2-2x+4}+\frac{317}{625}\int \frac{dx}{(x-1/5)^2+19/25}
=-\frac{x^2}{5}+11/25x+81/125\ln|5x^2-2x+4|+\frac{317}{625}\frac{1}{\sqrt{19}/5}\tan^{-1}(\frac{x-1/5}{\sqrt{19}/5})+C
=-\frac{x^2}{5}+11/25x+81/125\ln|5x^2-2x+4|+\frac{317}{125\sqrt{19}}\tan^{-1}(\frac{5x-1}{\sqrt{19}})+C
